Question

Find the equation of line parallel to $y$-axis and drawn through the point of intersection of the line $x-7y+5=0$ and $3x+y=0$.

• A. $x+22=0$
• B. $22x+5=0$
• C. $x+44=0$
• D. $22x-5=0$

Answer 1

Answer: (B)

Given, equation of lines are
$x-7y+5=0...(i)$
and $3x+y=0...(ii)$
On solving $(i)$ and $(ii)$, we get,
$x=\frac{-5}{22},y=\frac{15}{22}$
Hence, the intersection point is $\left(-\frac{5}{22},\frac{15}{22}\right)$.
$\therefore$ Equation of required line is
$y-\frac{15}{22}=\frac{1}{0}\left(x+\frac{5}{22}\right)$
$\Rightarrow 0=x+\frac{5}{22}$
$(\because$ Line is parallel to $y$-axis $\therefore m=tan90^{\circ }=\frac{1}{0})$
$\Rightarrow 22x+5=0$